Can Intersecting Events in Probability be Independent?

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Discussion Overview

The discussion centers on the independence of intersecting events in probability, specifically examining the conditions under which events A and B can be considered independent or dependent. Participants explore mathematical expressions and reasoning related to conditional probabilities and special cases in probability theory.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a mathematical derivation suggesting that if P(A∪B) = 1, then A and A∪B can be independent.
  • Another participant agrees that A and A∪B are independent under certain conditions but notes that there are other special cases to consider.
  • Concerns are raised about the implications of the intersection of A and B being empty, with a suggestion to refine the results by addressing exceptional cases.
  • One participant expresses a belief that intersecting events A and B must be dependent, questioning how to prove their dependence or independence.
  • Another participant counters that dependence or independence cannot be determined solely from the non-empty intersection of two events, emphasizing the need for additional assumptions or numerical values.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the conditions for independence of intersecting events. Multiple competing views are presented regarding the implications of the intersection and the necessary conditions for proving dependence or independence.

Contextual Notes

Participants highlight the need for additional assumptions or numerical values to fully address the questions of dependence and independence, indicating that the discussion is limited by the lack of specific probability values and definitions.

PFuser1232
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##P(A|A∩B) = \frac{P(A∩(A∩B))}{P(A∩B)} = \frac{P(A∩B)}{P(A∩B)} = 1##
So given the the event "A and B" as the sample space, the probability of A occurring is 1.

##P(A|A∪B) = \frac{P(A∩(A∪B))}{P(A∪B)} = \frac{P(A)}{P(A∪B)}##
Those two events are independent if and only if the probability of "A or B" occurring is 1, in which case the conditional probability of A equals the probability of A.

Is my reasoning correct?
 
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you mean that A and A∪B are independent if P(A∪B)=1 ? yes, this is a special case of independent events. There is one other special case which can mean that A and A∪B are independent. You are trying to make the last equation go from
P(A|A∪B) = \frac{P(A)}{P(A∪B)}
to become:
P(A|A∪B) = P(A)
right? and you did this by setting P(A∪B)=1. But there is another way also.
 
MohammedRady97 said:
##P(A|A∩B) = \frac{P(A∩(A∩B))}{P(A∩B)} = \frac{P(A∩B)}{P(A∩B)} = 1##
So given the the event "A and B" as the sample space, the probability of A occurring is 1.
We'd have to consider the case B \cap A = \emptyset

##P(A|A∪B) = \frac{P(A∩(A∪B))}{P(A∪B)} = \frac{P(A)}{P(A∪B)}##
Those two events are independent
Which two events? A and A \cup B ?

if and only if the probability of "A or B" occurring is 1

Suppose B = \emptyset.

I think your results are interesting and worth perfecting by taking care of the exceptional cases.
 
Stephen Tashi said:
We'd have to consider the case B \cap A = \emptysetWhich two events? A and A \cup B ?
Suppose B = \emptyset.

I think your results are interesting and worth perfecting by taking care of the exceptional cases.

What about ##P(A|B)##? My gut tells me that since A and B intersect, A and B must be dependent. This is wrong, of course, but how do I prove the dependence (or independence) of two intersecting events?
 
MohammedRady97 said:
What about ##P(A|B)##?

I don't know what your are asking.

My gut tells me that since A and B intersect, A and B must be dependent. This is wrong, of course, but how do I prove the dependence (or independence) of two intersecting events?

You can't prove the dependence or independence of two events that have a non-empty intersection just from knowing the intersection is non-empty. It's a quantitative question that depends on the numerical values of the probabilities involved. If you want to prove a result, you'll have to add more assumptions - something beyond just knowing that the intersection is non-empty.
 

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